
The equation for the combustion of glucose is: $ {C_6}{H_{12}}{O_6}(s) + 6{O_2} \to 6C{O_2}(g) + 6{H_2}O $ . How many grams of $ {H_2}O $ will be produced when $ 8.064g $ of glucose is burned?
Answer
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Hint :Combustion reaction: A type of chemical reaction in which a compound is heated in the presence of air to produce carbon dioxide and water. All combustion reactions are exothermic reactions as an appreciable amount of heat is released after the reaction.
Complete Step By Step Answer:
For finding mass of water produced, we first need to find out the number of moles of glucose burned. We know that, number of moles of any compound is equal to the ratio of mass used to the molar mass of the compound. Therefore, number of moles of glucose burned during the reaction are as follows:
$ n = \dfrac{{{\rm{given}}\,{\rm{mass}}}}{{{\rm{molar}}\,{\rm{mass}}}} $
Given mass of glucose i.e., mass of glucose burned during the reaction $ = 8.064g $
Molar mass of glucose $ = 180.16gmo{l^{ - 1}} $
$ \therefore $ number of moles of glucose $ n = \dfrac{{8.064}}{{{\rm{180}}{\rm{.16}}}} \Rightarrow 0.045{\rm{ moles}} $
The chemical reaction for combustion of glucose is given as follows:
$ {C_6}{H_{12}}{O_6}(s) + 6{O_2} \to 6C{O_2}(g) + 6{H_2}O $
As per the given reaction,
$ 1 $ mole of glucose reacts to produce $ \Rightarrow 6 $ moles of $ {H_2}O $
Therefore, $ 0.045 $ moles of glucose will react to produce $ \Rightarrow 6 \times 0.045 = 0.27 $ moles of $ {H_2}O $
Now, mass of water produced can be calculated using mole-mass relation which is as follows:
Number of moles of $ {H_2}O $ produced $ = \dfrac{{{\rm{mass}}}}{{{\rm{molar}}\,{\rm{mass}}}} $
Molar mass of $ {H_2}O = 18gmo{l^{ - 1}} $
Substituting values in the equation,
$ 0.27 = \dfrac{{mass}}{{18}} $
$ \Rightarrow mass = 18 \times 0.27 $
$ \Rightarrow mass = 4.86g $
Hence, $ 4.86g $ of $ {H_2}O $ is produced after the combustion of $ 8.064g $ glucose.
Note :
Combustion reaction is a type of oxidation reaction because of addition of oxygen atoms to the given compound. It is important to note that all combustion reactions are oxidation reactions but all oxidation reactions are not combustion reactions.
Complete Step By Step Answer:
For finding mass of water produced, we first need to find out the number of moles of glucose burned. We know that, number of moles of any compound is equal to the ratio of mass used to the molar mass of the compound. Therefore, number of moles of glucose burned during the reaction are as follows:
$ n = \dfrac{{{\rm{given}}\,{\rm{mass}}}}{{{\rm{molar}}\,{\rm{mass}}}} $
Given mass of glucose i.e., mass of glucose burned during the reaction $ = 8.064g $
Molar mass of glucose $ = 180.16gmo{l^{ - 1}} $
$ \therefore $ number of moles of glucose $ n = \dfrac{{8.064}}{{{\rm{180}}{\rm{.16}}}} \Rightarrow 0.045{\rm{ moles}} $
The chemical reaction for combustion of glucose is given as follows:
$ {C_6}{H_{12}}{O_6}(s) + 6{O_2} \to 6C{O_2}(g) + 6{H_2}O $
As per the given reaction,
$ 1 $ mole of glucose reacts to produce $ \Rightarrow 6 $ moles of $ {H_2}O $
Therefore, $ 0.045 $ moles of glucose will react to produce $ \Rightarrow 6 \times 0.045 = 0.27 $ moles of $ {H_2}O $
Now, mass of water produced can be calculated using mole-mass relation which is as follows:
Number of moles of $ {H_2}O $ produced $ = \dfrac{{{\rm{mass}}}}{{{\rm{molar}}\,{\rm{mass}}}} $
Molar mass of $ {H_2}O = 18gmo{l^{ - 1}} $
Substituting values in the equation,
$ 0.27 = \dfrac{{mass}}{{18}} $
$ \Rightarrow mass = 18 \times 0.27 $
$ \Rightarrow mass = 4.86g $
Hence, $ 4.86g $ of $ {H_2}O $ is produced after the combustion of $ 8.064g $ glucose.
Note :
Combustion reaction is a type of oxidation reaction because of addition of oxygen atoms to the given compound. It is important to note that all combustion reactions are oxidation reactions but all oxidation reactions are not combustion reactions.
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